Download Material and Energy Balances CHEN 2120 Outline Specific

Outline
Material and Energy Balances
CHEN 2120
Class Meeting #26
March 21st, 2007
– Reference states
“Energy balances on open systems”
Specific properties
• Specific property: an intensive quantity obtained
by dividing an extensive property (or its flow
rate) by the total amount (or flow rate) of the
process material.
• Extensive properties: mass, volume, kinetic
energy, potential energy, internal energy
• Intensive properties: specific mass, specific
volume, specific kinetic energy, etc….
• The symbol ^ denotes specific properties.
Open systems
inlet
• Specific properties and enthalpy
• Flow work and shaft work
• The steady-state open-system energy
balance
• Tables of thermodynamic data
• Steam tables
Specific properties
• Thus, when a specific property is multiplied by
mass or moles, the total extensive property can
be determined
• Example:
ˆ
U ( J / kg ) = m(kg )U ( J / kg )
Flow work and shaft work
outlet
• Material can flow into and out of an open
system
• Doesn’t necessarily have to be at steady
state (but we’ll do mostly steady state)
• W = Ws + Wfl
• Ws = shaft work, or rate of work done by
the process fluid on a moving part within
the system (e.g., pump or piston)
• Wfl = flow work, or rate of work done by
the fluid at the system outlet minus the
rate of work done on the fluid at the
system inlet
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Flow work
inlet
Enthalpy
outlet
• Wfl,in = - Pin·Vin (negative since work done on system)
• Wfl,out = Pout·Vout
• Wfl = PoutVout – PinVin
• Enthalpy: H = U + PV
• “Designed” to take into account ‘PV’ work for
open systems
• Example: If the specific internal energy of a gas
at 300 K and 1 atm is 4200 J/mol, and the
specific molar volume at the same temperature
and pressure is 25.34 L/mol, the specific
enthalpy of the gas is:
Ĥ
Ĥ
Enthalpy
•
•
•
•
∆U + ∆Ek + ∆Ep = Q – W
But W = Ws + Wfl
And Wfl = PoutVout – PinVin
Therefore,
Uout – Uin + Ek,out – Ek,in + Ep,out – Ep,in = Q – Ws – (PoutVout
– PinVin)
Re-arrange:
(Uout+PoutVout) – (Uin+PinVin) + Ek,out–Ek,in+Ep,out–Ep,in= Q–Ws
Æ ∆H + ∆Ek + ∆Ep = Q - Ws
• Specific Enthalpy:
Ĥ
= 4200 J/mol + (1 atm)(25.34 L/mol)(8.314 J/molK)/(0.08206 L-atm/K-mol)
= 6770 J/mol
Changes in enthalpy
•
Mainly when:
1) Heating or cooling a solid, liquid, or gas
2) Phase changes (evaporation, condensation,
freezing, melting)
∆H& + ∆E& k + ∆E& p = Q& − W& s
∆Hˆ + ∆Eˆ k + ∆Eˆ p = Qˆ − Wˆ s
Reference states
Reference states
• It is not possible to know the absolute
value of U or H for a process material, but
you can determine the change in U (∆U)
or in H (∆H) corresponding to a specified
change of state (temp., pressure, phase)
Mt. Everest
(29,029 ft)
Denali
(20,320 ft)
Bear Peak
(8,250 ft)
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Reference states
Mt. Everest
(29,029 ft)
Denali
(20,320 ft)
Bear Peak
(8,250 ft)
Now you try… (Clicker Prob. 26.1)
How tall is Bear Peak?
(depends upon ref!)
How much enthalpy does
stream X have?
(depends upon ref!)
How much taller is Mt.
Everest than Denali
(does not depend upon
ref, but you do need a ref. state)
What is the change in
enthalpy going from state
A to B?
(does not depend upon ref, but
you do need a ref. state)
Mariana Trench
(-35,000 ft)
Now you try… (Clicker Prob. 26.1)
State
T(K)
P(bar)
V’(L/mol) U’(kJ/mol)
Liquid
300
0.310
0.0516
0.000
Vapor
300
0.310
79.94
28.24
Vapor
340
1.33
20.92
29.62
•
State
T(K)
P(bar)
V’(L/mol) U’(kJ/mol)
Liquid
300
0.310
0.0516
0.000
Vapor
300
0.310
79.94
28.24
Vapor
340
1.33
20.92
29.62
•
What reference state was used to generate the
listed specific internal energies?
A) Liquid at 300 K
B) Vapor at 300 K
C) Vapor at 340 K
Steam Tables
Calculate ∆U’(kJ/mol) for a process in which
bromine vapor at 300 K is condensed at
constant pressure. Then calculate ∆H’(kJ/mol)
for the same process.
Example
• Determine the vapor pressure, specific
internal energy, and specific enthalpy of
saturated steam at 133.5°C.
Example
• Determine the vapor pressure, specific
internal energy, and specific enthalpy of
saturated steam at 133.5°C.
• From Table B.6: P = 3.0 bar, U’ = 2543.0
kJ/mol, H’ = 2724.7 kJ/mol
3
Now you try… (Clicker Prob. 26.2)
inlet
outlet
• Steam enters an isobaric heat exchanger at a mass flow
rate of 0.08 kg/sec as a superheated vapor at 350°C and
1.0 bar and exits the heat exchanger as a superheated
vapor at 150°C. How much heat (kJ/s) is transferred
from the steam to the heat exchanger?
A) 2.4 kJ/s
B) 45.2 kJ/s
C) 32.0 kJ/s
D) 10.4 kJ/s
Now you try… (Clicker Prob. 26.2)
inlet
outlet
• H’in = 3176 kJ/kg
• H’out = 2776 kJ/kg
• Q = ∆H = (0.08 kg/s)(3176 – 2776)kJ/kg
= 32 kJ/s (or 32 kW) (Ans. C)
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